Përplasje plotësisht elastike
Një përplasje e dy objekteve quhet përplasje plotësisht elastike nëse impulsi ose energjia kinetike e secilit objekt para përplasjes është e barabartë me impulsin dhe energjinë kinetike të secilit objekt pas përplasjes. Me fjalë të tjera, ligji i ruajtjes së impulsit dhe ligji i ruajtjes së energjisë kinetike janë të zbatueshëm në përplasjet plotësisht elastike. Përdorimi i fjalës elastik tregon se pas përplasjes, dy objektet nuk ngjiten së bashku ose nuk janë të bashkangjitura me njëri-tjetrin, por kërcejnë. Impulsi i secilit objekt ruhet.
Impulsi i secilit objekt ruhet.
m1 v1 +m2 v2 = m1 v1 ' + m2 v2 Ekuacioni 1.5
The kinetic energy of each object is conserved.
1⁄2 mV12 + 1⁄2 m v22 = 1⁄2 m v1'2 + 1⁄2 m v2'2 ………………….. Equation 1.6
The perfectly elastic collision must be silent and does not generate heat due to friction between the two colliding objects. If the collision of two objects generates noise and heat, the kinetic energy of the objects is not conserved. Some kinetic energy is converted into sound energy and heat energy, and some are converted into internal energy. An example of a perfectly elastic collision is the collision of atomic and subatomic particles.
In a problem of perfectly elastic collision, if the initial speed is known while the final speed is unknown, the problem cannot be solved by only using equations 1.5 and 1.6. For this reason, both of the equations are manipulated to derive other equations, which can be used to determine the final speed.
Remove factor 1/2 then manipulate 1.6
m1 v12 +m2 v22 = m1 v1 ' 2 +m2 v2 ' 2
m1 v12 - m1 v1' 2 = m2 v2 ' 2 - m2 v22
m1 (v12 - v1' 2 ) = m2 (v2 ' 2 - v2 2) —> (a + b)(a – b) = a2 - b2
m1 (v1 +v1 ’) (v1 - v1 ’) = m2 (v2 ' + v2 ) (v2 ’ – v2 ) ………………….. Equation 1.7
Manipulate equation 1.5
m1 v1 - m1 v1 ’ = m2 v2 ’- m2 v2
m1 (v1 - v1 ’) = m2 (v2 ’ – v2) ………………….. Equation 1.8
Devide equation 1.7 by equation 1.8 to obtain the final result
(v1 +v1 ’) = (v2 ' + v2)
v1 - v2 v2 ’ – v1 '
v1 - v2 = – (v1 ’ – v2 ’) ………………….. Equation 1.9
Equations 1.5 and 1.9 can be used to solve problems of perfectly elastic collisions. Combine equations 1.5 and 1.9 to obtain two equations for determining the final velocities of two objects if their masses and initial velocities are unknown.
When solving questions using the equations above, use the correct sign for v1 dhe v2. If object 1 moves to the right, v1 is positive, and conversely, if object two moves to the left, v2 is negative. If both objects move in different directions, but there is no information regarding the motion directions, v1 dhe v2 must be signed differently, for example, v1 is positive, and v2 është negativ.
4.1.1 Two objects of the same mass
If m1 = m2, v1 ' = v2 dhe v2' = v1 .
What if the two objects move in different directions?
For instance, if before collision object 1 moves to the right and object 2 moves to the left, object 1 will move to the left (v1' = – v2) and object 2 will move to the right (v2' = v1) after the collision.
What if either of the objects is initially at rest?
For instance, if before collision object 2 is at rest (v2 = 0), object 1 will be at rest (v1’ = 0) and object 2 will move at the same velocity as the initial velocity of object 1 (v2' = v1) after collision. If before collision object 1 moves to the right, object 2 will move to the right after the collision. Hence, the two objects exchange velocities.
Shembull pyetjeje 4
Objects A (2 kg) and B (2 kg) move to opposite directions at speeds of 4 m/s and 2 m/s, respectively. If objects A and B collide in a perfectly elastic collision, what are the final speeds of objects A and B?
I njohur:
mA = 2 kg, mB = 2 kg, vA = 4 m/s, vB = – 2 m/s
Kërkohet: vA’ and vB '
zgjidhje:
vA’ = – 2 m/s and vB' = 4 m/s
After the collision, object A moves at a speed of 2 m/s and object B moves at a speed of 4 m/s in opposite directions. If before collision object A moves to the right and object B moves to the left, object A will move to the left and object B will move to the right after the collision.
Shembull pyetjeje 5
Object A (2 kg) moves to the right at a speed of 2 m/s and collides with object B (2 kg) which is at rest. If the two objects collide in a perfectly elastic collision, what are the final speeds of objects A and B?
I njohur:
mA = 2 kg, mB = 2 kg, vA = 2 m/s, vB = 0 m/s
Kërkohet: vA ’ and vB '
zgjidhje:
vA’ = 0 and vB' = 2 m/s
After the collision, object A is at rest, and object B moves to the right at a speed of 2 m/s.
4.1.2 Two objects of different masses
If two objects have different initial velocities and masses (minor difference), the final velocities are known by using equations 1.10 and 1.11.
If initially object 2 is at rest (v2 = 0), equation 1.10 becomes equation 1.12 and equation 1.11 becomes equation 1.13.
Nëse v2 = 0, m1 is very great, and m2 is very small, by solving equations 1.12 and 1.13 v1' = v1 dhe v2' = 2v1 are obtained. If v2 = 0, m1 is very small, and m2 is very great, by solving equations 1.12 and 1.13 v1 ' = -v1 dhe v2 ’ = 0 are obtained. Positive and negative marks indicate opposite motion directions.
Shembull pyetjeje 6
An object with a mass of 1 kg moves at a speed of 20 m/s and collides with a wall in a perfectly elastic collision. What are the final speeds of the object and the wall?
I njohur:
mA = 1 kg, vA = 20 m/s, mB = very great, vB = 0
Shtepi :vA’ and vB'
zgjidhje:
vA’ = -20 m/s and vB' = 0
After collision, object A bounces off at a speed of 20 m/s and object B remains at rest. If before collision object A moves to the right, it moves to the left after collision.
